Discussion
A spherical ball weighing 2 kg is dropped from a height of 4.9 m onto an immovable rigid block as shown in the figure. If the collision is perfectly elastic, what is the momentum vector of the ball (in kg m/s) just after impact?
Take the acceleration due to gravity to be \(g = 9.8\) m/s2. Options have been rounded off to one decimal place.
Answer: C
The momentum just after impact is \(17.0\,\hat{\imath} + 9.8\,\hat{\jmath}\) kg m/s.
Speed at impact and momentum, pointing straight down:
\[ \begin{aligned} v &= \sqrt{2gh} = \sqrt{2 \times 9.8 \times 4.9} \\ &= 9.8\ \text{m/s} \\ \vec{p} &= 19.6\ \text{kg m/s, downward} \end{aligned} \]
In a perfectly elastic impact with a fixed smooth surface, the component of momentum along the surface is kept and the component along the normal is reversed. The face falls to the right at 30°, so its outward normal is \(\hat{n} = \sin 30^{\circ}\,\hat{\imath} + \cos 30^{\circ}\,\hat{\jmath}\). Then
\[ \begin{aligned} \vec{p}\cdot\hat{n} &= -19.6 \times 0.866 = -16.97 \\ \vec{p}' &= \vec{p} - 2(\vec{p}\cdot\hat{n})\,\hat{n} \\ &= -19.6\,\hat{\jmath} \\ &\quad + 33.95(0.5\,\hat{\imath} + 0.866\,\hat{\jmath}) \\ &= 16.97\,\hat{\imath} + 9.8\,\hat{\jmath} \end{aligned} \]
The size stays 19.6, as it must when no energy is lost (\(16.97^{2} + 9.8^{2} = 19.6^{2}\)). \(19.6\,\hat{\jmath}\) is the rebound from a horizontal floor, and \(19.6\,\hat{\imath}\) a deflection through 90°. \(9.8\,\hat{\imath} + 17.0\,\hat{\jmath}\) swaps the components, which is what a 60° face would give.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development